PROCEEDINGS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 129, Number 6, Pages 1587–1593 S 0002-9939(00)05809-3 Article electronically published on October 31, 2000

PRIMITIVE NOETHERIAN ALGEBRAS WITH BIG CENTERS

RONALD S. IRVING

(Communicated by Ken Goodearl)

Abstract. Recent work of Artin, Small, and Zhang extends Grothendieck’s classical commutative algebra result on generic freeness to a large family of non-commutative algebras. Over such an algebra, any finitely-generated mod- ule becomes free after localization at a suitable central element. In this paper, a construction is given of primitive noetherian algebras, finitely generated over the integers or over algebraic closures of finite fields, such that the faithful, simple modules don’t satisfy such a freeness condition. These algebras also fail to satisfy a non-commutative version of the Nullstellensatz.

1. Introduction

Given a commutative R and a non-zero element s of R,letRs denote the localization R[s−1]. An R- M is generically free if there is a non-zero s in R such that Rs ⊗R M is free over Rs. As a trivial example, if R is a G-domain, a domain whose fraction field has the form Rs for some non-zero s,theneveryR- module is generically free. More importantly, Grothendieck proved for R noetherian and A a finitely-generated commutative R-algebra that any finitely-generated A- module is generically free over R. For a non-commutative R-algebra A, work of Quillen, Dixmier, and Duflo related generic freeness of A-modules over of the center of A to the study of primitive ideals of A. Suppose A is a primitive with a faithful, M. By Schur’s Lemma, the EndA(M) is a , and the center Z of A embeds in EndA(M). Hence any R of Z is a domain whose fraction field lies in EndA(M). If R is a field, M is free over R,andifR is a G-domain, M is generically free over R.Conversely,ifM is free over R,ashort argument shows that R is a field, and if M is generically free over R, a second short argument shows that R is a G-domain. (See Proposition 1 of [3] or Proposition 4 of [4].) Thus, given a subring R of Z that is not a G-domain, such as Z or a polynomial ring k[t] over a field, the faithful, simple A-modules are not generically free over R. New interest in generic freeness has arisen from recent work of Artin, Small, and Zhang [1] extending the generic freeness theorem of Grothendieck to a broad family of non-commutative algebras. The preceding discussion shows in contrast that primitive rings with big centers have simple modules for which generic freeness does not hold. In [4], a finitely-generated, primitive, noetherian Z-algebra A was

Received by the editors September 9, 1999. 2000 Mathematics Subject Classification. Primary 16D60, 16P40; Secondary 16S36. Key words and phrases. Generic freeness, Nullstellensatz.

c 2000 American Mathematical Society 1587

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constructed; since Z is not a G-domain, the faithful, simple A-modules are not generically free over Z. For certain fields k, a finitely-generated, noetherian k[t]- algebra was constructed in [5] and primitivity of the algebra was asserted. However, the proof of primitivity is incorrect. A general construction of primitive rings will be described in Section 2, and used in Section 3 to produce as special cases the example in [4] and a corrected version of the example in [5]. The following theorem is a consequence.

Theorem 1.1. Let p be a prime number and let kp be an algebraic closure of the finite field of p elements. There is a finitely-generated, primitive, noetherian kp- algebra Ap containing a polynomial ring kp[t] in its center, and the faithful, simple Ap-modules are not generically free over kp[t]. The original interest in the connection between primitivity and generic freeness flowed in the opposite direction, with generic freeness of algebras in certain families being used to restrict the possibilities for their primitive ideals. An algebra A over a field k satisfies the Nullstellensatz if the endomorphism ring of any simple A-module M is algebraic over k.Fork algebraically closed, A satisfies the Nullstellensatz precisely if the endomorphism ring of any simple A-module is k itself. For the poly- nomial ring k[x1,...,xn], this condition is a version of the classical Nullstellensatz of Hilbert. Suppose A is a k-algebra such that any finitely-generated A[t]-module is generically free over k[t]. The following simple argument shows that A satisfies the Nullstellensatz. Given a simple A-module M,letE be an A-endomorphism, regard M as an A[t]-module with t acting like E,andletP = AnnA[t]M.IfP ∩ k[t]were zero, then k[t] would embed in the primitive ring A[t]/P and the generic freeness of M over k[t] would imply that k[t]isaG-domain. Hence P ∩ k[t] must be non-zero, and E is algebraic over k. In 1963, Dixmier proved by an elementary counting argument that any finitely- generated algebra over a field k of uncountable cardinality satisfies the Nullstel- lensatz [2]. His motivation lay in the study of enveloping algebras of Lie algebras, and this result was fundamental in the analysis of their primitive ideals. The associated graded algebra of an enveloping algebra is a polynomial ring, to which Grothendieck’s generic freeness theorem can be applied. In a 1969 paper [6], Quillen used generic freeness to prove that the Nullstellensatz holds for enveloping algebras over any base field. In 1973, following an argument of Dixmier, Duflo proved for an enveloping algebra B over a commutative domain S that finitely-generated B- modules are generically free over S [3], and deduced the Nullstellensatz anew by the argument above. The algebras in [5] were intended to be examples of finitely-generated noetherian algebras that fail to satisfy the Nullstellensatz. Theorem 1.1 provides just such examples — since the endomorphism ring of any faithful, simple Ap-module contains kp(t), it is not algebraic over kp. The theorem is proved by constructing the algebra Ap and an explicit faithful, simple Ap-module whose endomorphism ring is precisely kp(t). I thank Michael Artin and James Zhang for bringing the error in [5] to my attention, and for suggesting that a mild variation should still work.

2. The construction Let K be a field, let a and b be non-zero elements of K,andletφ : K[y] → K[y] be the ring automorphism that fixes K identically and sends y to ay + b.The inverse of φ is given by φ−1(y)=a−1y − a−1b.LetB be the twisted Laurent

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extension K[y][x, x−1; φ], consisting of finite K[y]-linear combinations of positive and negative powers of x. Addition is defined in the obvious way and multiplication is given by the rule x−1g(y)x = φ(g(y)) for g(y)inK[y]. In particular, elements of K commute with powers of x,and −1 n x yx = ay + b. For each integer n,letrn be the element φ (y)inK[y]. Then for n positive, n n−1 rn = a y +(a + ···+ a +1)b and −n −n −1 r−n = a y − (a + ···+ a )b. The homomorphism π : K[y] → K that fixes K identically and sends y to 1 defines a K[y]-module structure on K. For each integer n,letcn = π(rn). Then c0 =1 and for n positive, n n−1 cn = a +(a + ···+ a +1)b and −n −n −1 c−n = a − (a + ···+ a )b.

Let V be the B-module B ⊗K[y] K obtained from the K[y]-module K by extension of scalars.

Proposition 2.1. Suppose that the set {cn : n ∈ Z} of constants in the field K is infinite. Then the B-module V is faithful and simple. Proof. The ring B is free as a K[y]-module, with a basis consisting of the powers of x. Therefore, as a K-, V has the basis {xn ⊗ 1:n ∈ Z}. n Let vn denote the basis vector x ⊗ 1. The action of x on this basis is given by x.vn = vn+1. The assumption that π(y) = 1 yields y.v0 = v0.Forn a non-zero integer, n n −n n n n y.vn = yx .v0 = x x yx .v0 = x rn.v0 = x cn.v0 = cn.vn.

Thus each basis vector vn is a y-eigenvector, with eigenvalue cn. To prove that V is simple, let v be a non-zero element of V . We will prove that B.v contains a non-zero K-multiple of some vn.Ifv is already of this form, there is nothing to do. Otherwise, v has the form

v = αivi + αi+1vi+1 + ···+ αi+jvi+j ,

with j positive and αiαi+j =6 0. Since the set {cn} of y-eigenvalues is infinite, there t−i is an index t such that ct =6 ct+j. As a result, the vector (y −ct)x .v, which equals

αi(ct − ct)vt + αi+1(ct+1 − ct)vt+1 + ···+ αi+j(ct+j − ct)vt+j , is a non-zero element of B.v that involves fewer basis vectors than v does. By induction, B.v contains a non-zero K-multiple of some vn, as desired. Using the actions of K and the powers of x on this vector, we obtain B.v = V ,andV is simple. For faithfulness, suppose an element b of B annihilates V .Writingb as a K[y]- linear combination of powers of x and applying this to each of the basis vectors

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of V , we see that each coefficient of b in K[y] also annihilates V . Thus it suffices to show that if a polynomial g(y)inK[y] annihilates V ,theng(y)=0.But g(y).vn = g(cn).vn.Thusg(y) annihilates V if and only if g(y)vanishesonevery cn, and this is only possible if g(y)=0.

If the constants cn are all non-zero, then a localization of B can also be realized asaringofoperatorsonV .LetT be the multiplicatively closed subset of K[y] consisting of finite products of the elements {rn : n ∈ Z},andletS = K[y]T . −1 Since r0 = y, the ring S contains y . The automorphism φ of K[y] extends to an n −1 n −1 −1 automorphism of S, also denoted φ,withφ (y )=(φ (y)) = rn .LetC be the twisted Laurent extension S[x, x−1; φ], containing B as a K-subalgebra. Suppose the constants cn in K are all non-zero. Then the homomorphism π : → −1 −1 K[y] K extends to S,withπ(rn )=cn .ThismakesK an S-module, and extending scalars yields the C-module C ⊗S K.JustasB is free as a K[y]-module, with basis the powers of x,sotooC is free as an S-module, with the same basis. Therefore, as in the proof of Proposition 2.1, the module C ⊗S K has the basis {xn ⊗ 1:n ∈ Z}

as a K-vector space. This allows us to identify C ⊗S K with V .TheC-module structure on V induces a map from C to EndK (V ), and the image of C in EndK (V ) contains an isomorphic copy of B. Therefore V is simple as a C-module. The argument in the proof of Proposition 2.1 can be used to show that V is also C- faithful — it suffices to show that any element s of S annihilating V is 0. Clearing denominators of such an s yields an element of K[y] annihilating V ,andthismust be 0. Hence:

Proposition 2.2. Suppose that the set {cn : n ∈ Z} of constants in the field K is infinite and excludes 0. Then the C-module V is faithful and simple. Suppose that the field K is the fraction field of a domain Q, that the elements a and b of K lie in Q,andthata is a unit in Q. Then the automorphism φ of K[y] restricts to an automorphism of Q[y], again denoted φ. The elements rn all lie in Q[y] and the constants cn all lie in Q. Therefore T is a multiplicatively closed subset of Q[y]andφ extends to an automorphism of Q[y]T .LetR = Q[y]T and let A be the twisted Laurent extension R[x, x−1; φ]. Since the ring C of Proposition 2.2 contains A as a subring, the C-module V becomes an A-module under restriction of scalars.

Proposition 2.3. Suppose that the subset {cn : n ∈ Z} of constants in the domain { −1 ∈ Z} Q is infinite and excludes 0, and suppose further that the set cn : n generates the fraction field K as a Q-algebra. Then the A-module V is faithful and simple with endomorphism ring K. Proof. Since V is faithful over C, it certainly is over A. Given a non-zero element v of V , the actions of y and powers of x can be used, as in the proof of Proposition 2.1, to show that there is a non-zero α in K with αv0 ∈ A.v. For any integer n, −n −1 n −1 x y x .v0 = cn .v0. { −1} Since the set cn generates K over Q, the submodule A.v0 contains K.v0.But then A.(αv0) also contains K.v0,andA.v = V ,provingthatV is simple.

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Suppose E is an abelian group endomorphism of V that commutes with the action of A.Let

Xt E(v0)= αivi i=s

for scalars αi in K and indices s ≤ t. For any integer n,

Xt n n E(vn)=E(x .v0)=x .E(v0)= αivi+n. i=s Hence

Xt Xt cn αivi+n = cnE(vn)=E(cn.vn)=E(y.vn)=y.E(vn)= ci+nαivi+n. i=s i=s Subtracting the right side from the left yields

Xt (cn − ci+n)αivi+n =0. i=s

Given an index i with αi =6 0, the equality cn = ci+n holds for all integers n.This can only occur if i =0.ThusE(v0)=α0v0 and E acts on all of V as scalar multiplication by α0.

The following special case of Proposition 2.3 will be used in Section 3.

Corollary 2.4. Let P be a noetherian domain with fraction field K,letd be a non- zero element of P ,andletQ = P [d−1].Leta be a unit in Q,letb be a non-zero element of Q,letφ : Q[y] → Q[y] be the automorphism that sends y to ay + b, and let T be the multiplicatively closed subset of Q[y] consisting of finite products n of the elements {φ (y):n ∈ Z}.Extendφ to an automorphism of Q[y]T and let A −1 be the twisted Laurent extension Q[y]T [x, x ; φ]. For each integer n, assume that n the evaluation cn of φ (y) at y =1is non-zero. Assume in addition that the set {cn : n ∈ Z} is infinite and that every irreducible element of P divides either d or some cn. • A is a finitely-generated, noetherian P -algebra. • A is primitive, with a faithful, simple module V whose endomorphism ring is K.

Proof. As a P -algebra, A is generated by the set

{d−1,y,y−1,x,x−1}.

Since P is noetherian, so is Q[y]T , and the usual Hilbert Basis Theorem argument shows that A is noetherian as well. Also since P is noetherian, every non-zero non-unit of P is a finite product of irreducible elements. Hence the fraction field K of P is generated as a P -algebra by the set of inverses of the irreducible elements of P , ensuring that the hypotheses of Proposition 2.3 hold.

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3. Examples In this section we will apply Corollary 2.4 to two families of examples. For the first family, let P be the ring of integers and let K be the field of rational numbers. A natural choice for an automorphism of Q[y] is the map that sends y to y +1. However, with this choice, cn = n +1,sothat c−1 = 0 and Corollary 2.4 cannot be applied. To avoid this difficulty, we use a higher power. Example 3.1. Let d be an integer greater than 1 and let φ : Z[y] → Z[y] be the ring automorphism defined by φ(y)=y + d.LetT be the multiplicatively closed subset of Z[y] consisting of finite products of the polynomials in the set {y + dn : n ∈ Z}

−1 and let R = Z[d ][y]T .Extendφ to R and let A be the twisted Laurent extension R[x, x−1; φ]. • A is a finitely-generated, primitive, noetherian Z-algebra with a faithful, sim- ple module V whose endomorphism ring is Q. • The faithful, simple A-modules are not generically free over Z.

Proof. Each constant cn is dn + 1. Given a prime number p that doesn’t divide d, the Euclidean algorithm yields integers m and n for which pm + dn = −1. Hence p divides dn + 1 and the hypotheses of Corollary 2.4 are satisfied. Given a faithful, simple A-module W , the discussion in the second paragraph of Section 1 shows that since AnnZW is zero, W cannot be generically free over Z, and this completes the proof.

If d = 2 in Example 3.1, then the constants cn form the set of odd numbers and the verification that the hypotheses of Corollary 2.4 are satisfied reduces to the observation that every prime number divides either 2 or an odd number. The resulting Z-algebra is the one appearing in [4]. For the second family of examples, let P be the polynomial ring k[t]overafieldk and let K be the field k(t) of rational functions. The automorphism of k(t)[y]that sends y to ty + 1, like the automorphism of Q[y] that sends y to y +1,has0 asthe constant c−1, so that Corollary 2.4 doesn’t apply. This difficulty can be avoided, just as in Example 3.1, if a higher power of the automorphism is used. Then, in order for the assumption in Corollary 2.4 on divisibility by irreducible elements to be satisfied, a careful choice of the base field k must be made. Example 3.2. Let p be a prime number and let k be an algebraic closure of the finite field Fp of p elements. Let φ : k(t)[y] → k(t)[y] be the ring automorphism that is the identity on k(t) and that is defined on the rest of k(t)[y] by the rule φ(y)=tpy + tp−1 + tp−2 + ···+ t +1. Let T be the multiplicatively closed subset of k[t][y] consisting of finite products of the polynomials in the set {y}∪{tnpy +(tnp−1 + tnp−2 + ···+ t +1):n ∈ Z+} ∪{t−npy − (t−np + t−np+1 + ···+ t−1):n ∈ Z+}.

2 −1 Let d = t − t and let R = k[t][d ][y]T .Extendφ to R and let A be the twisted Laurent extension R[x, x−1; φ].

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• A is a finitely-generated, primitive, noetherian k-algebra with a faithful, simple module V whose endomorphism ring is k(t). • The faithful, simple A-modules are not generically free over k[t]. • A does not satisfy the Nullstellensatz. Proof. The set {t, d−1,y,y−1,x,x−1} generates A as a k-algebra. Suppose q(t)is an irreducible polynomial of k[t] that doesn’t divide d. Then there is an element α of k distinct from 0 or 1 such that q(t) is a non-zero k-multiple of t − α.Let m m be the positive integer such that the field Fp[α] has cardinality p ,andlet m n = pm − pm−1 − 1. Then αp −1 =1and m αnp+1 =(αp −1)p−1 =1. Hence t − α divides tnp+1 − 1. But α =1,so6 t − α also divides the quotient (tnp+1 − 1)/(t − 1), which is tnp + ···+ t + 1. This last polynomial is precisely cn, and we have proved that every irreducible polynomial of k[t] divides either d or some cn. Corollary 2.4 and the discussion in Section 1 complete the proof. References

1. M. Artin, L. W. Small, and J. J. Zhang, Generic flatness for strongly noetherian algebras,J. Algebra 221 (1999), 579–610. CMP 2000:05 2. J. Dixmier, Repr´esentations irr´eductibles des alg`ebres de Lie nilpotents, An. Acad. Brasil. Ciˆenc. 35 (1963), 491–519. MR 32:165 3. M. Duflo, Certaines alg`ebres de type fini sont alg`ebres de Jacobson,J.Algebra27 (1973), 358–365. MR 49:9018 4. R. S. Irving, Generic flatness and the Nullstellensatz for Ore extensions, Comm. Algebra 7 (1979), 259–277. MR 80b:13005 5. R. S. Irving, Noetherian algebras and the Nullstellensatz,inS´eminaire d’Alg`ebre Paul Dubreil, Proceedings, Paris 1977-1978, Springer-Verlag, Berlin, 1979, pp. 80-87. MR 81c:16018 6. D. Quillen, On the endomorphism ring of a simple module over an enveloping algebra,Proc. Amer. Math. Soc. 21 (1969), 171–172. MR 39:252

Department of Mathematics, University of Washington, Seattle, Washington 98195 E-mail address: [email protected]

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